how to solve PHP Parse error: syntax error, unexpected token "echo" in php shell code on line 2 in interactive

I opened an interactive PHP shell using php -a and started entering the statements

php > "12" == 12
php > echo "12" == 12
php > echo ("12" == 12)
php > echo ("12" == 12);

And then I got the error

PHP Parse error:  syntax error, unexpected token "echo" in php shell code on line 2

I was expecting to get true or false, why doesn't the interactive shell tell me what the value of "12" = 12 is?

I opened an interactive PHP shell using php -a and started entering the statements

php > "12" == 12
php > echo "12" == 12
php > echo ("12" == 12)
php > echo ("12" == 12);

And then I got the error

PHP Parse error:  syntax error, unexpected token "echo" in php shell code on line 2

I was expecting to get true or false, why doesn't the interactive shell tell me what the value of "12" = 12 is?

• php
• interactive-shell

Share Improve this question Follow asked Feb 3 at 12:59 camilajennycamilajenny 5,09466 gold badges3535 silver badges6767 bronze badges

Add a comment  | 

1 Answer 1

Sorted by: Reset to default 2

The error was because I didn't finish the instructions, so the interactive shell interpreted my statements as one long instruction

"12" == 12 echo "12" == 12 echo ("12" == 12) echo ("12" == 12);

which has a syntax error on the first echo.

The solution is to end the instruction with a semicolon

 echo "12" == 12 ? 'true' : 'false';